Question

Find the slope of the normal line to y=x+cos(xy) at (0,1)

Answer 1

Answer with explanation:

The equation of the curve is:  

         y= x + Cos ( x y)

Differentiating once to get,slope of tangent

y'=1 + Sin (x y)(y + x y')

Now, slope of tangent of the curve,at (0,1) can be obtained by,substituting,x=0 and y=1 , in the equation of the slope of tangent to the curve

$y'_{(0,1)}=1 + Sin (0 \times 1)(1+0 \times y'_{(0,1)})\\\\ \text{as},Sin 0^{\circ}=0\\\\y'_{(0,1)}=1$

Slope of tangent = 1

For, any curve,the two lines passing through , (0,1),that is normal line and tangent line will be perpendicular to each other,so as their slopes.

So,if two lines are perpendicular to each other

Product of their Slopes = -1

So, →Slope of tangent × Slope of Normal = -1

   → 1 ×Slope of Normal=-1

→Slope of Normal= -1

Equation of line passing through, (0,1) and having slope=-1,that is equation of normal line is

→y-1= -1×(x-0)

→ y-1=-x

→ x + y-1=0→→→Equation of normal line

Slope of normal line = -1

 

 

Answer 2

Derivating the equation:

$y=x+\cos(xy)\Longrightarrow y'=1+(-y\sin(xy)-xy'\sin(xy))\iff \\\\y'=1-y\sin(xy)-xy'\sin(xy)\iff y'+xy'\sin(xy)=1-y\sin(xy)\iff \\\\y'=\dfrac{1-y\sin(xy)}{1+x\sin(xy)}$

So the slope of the tangent in the point (0,1) is:

$y'=\dfrac{1-y\sin(xy)}{1+x\sin(xy)}\Longrightarrow y'=\dfrac{1-1\sin(0\cdot1)}{1+0\sin(0\cdot1)}\Longrightarrow y'=1$

Then, the slope of the normal (n) line in the point (0,1) is:

$n\cdot y'=-1\Longrightarrow n\cdot1=-1\iff\boxed{n=-1}$